https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_5&feed=atom&action=history
1969 Canadian MO Problems/Problem 5 - Revision history
2024-03-29T01:47:48Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_5&diff=27889&oldid=prev
1=2: /* Solution */
2008-09-10T16:35:33Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:35, 10 September 2008</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because <math>\sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>2ab\cos \frac C2=d(a+b).</math> Dividing by <math>a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Because <math>\sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>2ab\cos \frac C2=d(a+b).</math> Dividing by <math>a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>{{Old CanadaMO box|num-b=<del class="diffchange diffchange-inline">1</del>|num-a=<del class="diffchange diffchange-inline">3</del>|year=1969}}</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>{{Old CanadaMO box|num-b=<ins class="diffchange diffchange-inline">4</ins>|num-a=<ins class="diffchange diffchange-inline">6</ins>|year=1969}}</div></td></tr>
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1=2
https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_5&diff=19642&oldid=prev
Temperal: box
2007-11-18T02:40:45Z
<p>box</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:40, 18 November 2007</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let <math><del class="diffchange diffchange-inline">\displaystyle </del>ABC</math> be a triangle with sides of length <math><del class="diffchange diffchange-inline">\displaystyle </del>a</math>, <math><del class="diffchange diffchange-inline">\displaystyle </del>b</math> and <math><del class="diffchange diffchange-inline">\displaystyle </del>c</math>. Let the bisector of the <math><del class="diffchange diffchange-inline">\displaystyle </del>\angle C</math> cut <math><del class="diffchange diffchange-inline">\displaystyle </del>AB</math> at <math><del class="diffchange diffchange-inline">\displaystyle </del>D</math>. Prove that the length of <math><del class="diffchange diffchange-inline">\displaystyle </del>CD</math> is <math><del class="diffchange diffchange-inline">\displaystyle </del>\frac{2ab\cos \frac{C}{2}}{a+b}.</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let <math>ABC</math> be a triangle with sides of length <math>a</math>, <math>b</math> and <math>c</math>. Let the bisector of the <math>\angle C</math> cut <math>AB</math> at <math>D</math>. Prove that the length of <math>CD</math> is <math>\frac{2ab\cos \frac{C}{2}}{a+b}.</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let <math><del class="diffchange diffchange-inline">\displaystyle </del>CD=d.</math> Note that <math><del class="diffchange diffchange-inline">\displaystyle </del>[\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math><del class="diffchange diffchange-inline">\displaystyle </del>\frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let <math>CD=d.</math> Note that <math>[\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math>\frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Because <math><del class="diffchange diffchange-inline">\displaystyle </del>\sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math><del class="diffchange diffchange-inline">\displaystyle2ab</del>\cos \frac C2=d(a+b).</math> Dividing by <math><del class="diffchange diffchange-inline">\displaystyle </del>a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Because <math>\sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math><ins class="diffchange diffchange-inline">2ab</ins>\cos \frac C2=d(a+b).</math> Dividing by <math>a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>--<del class="diffchange diffchange-inline">--</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">{{Old CanadaMO box|num</ins>-<ins class="diffchange diffchange-inline">b=1|num</ins>-<ins class="diffchange diffchange-inline">a=3</ins>|<ins class="diffchange diffchange-inline">year=</ins>1969<ins class="diffchange diffchange-inline">}}</ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">* [[1969 Canadian MO Problems/Problem 4|Previous Problem]]</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">* [[1969 Canadian MO Problems/Problem 6</del>|<del class="diffchange diffchange-inline">Next Problem]]</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">* [[</del>1969 <del class="diffchange diffchange-inline">Canadian MO Problems|Back to Exam]]</del></div></td><td colspan="2"> </td></tr>
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Temperal
https://artofproblemsolving.com/wiki/index.php?title=1969_Canadian_MO_Problems/Problem_5&diff=8604&oldid=prev
4everwise at 06:04, 28 July 2006
2006-07-28T06:04:08Z
<p></p>
<p><b>New page</b></p><div>== Problem ==<br />
Let <math>\displaystyle ABC</math> be a triangle with sides of length <math>\displaystyle a</math>, <math>\displaystyle b</math> and <math>\displaystyle c</math>. Let the bisector of the <math>\displaystyle \angle C</math> cut <math>\displaystyle AB</math> at <math>\displaystyle D</math>. Prove that the length of <math>\displaystyle CD</math> is <math>\displaystyle \frac{2ab\cos \frac{C}{2}}{a+b}.</math><br />
<br />
<br />
== Solution ==<br />
Let <math>\displaystyle CD=d.</math> Note that <math>\displaystyle [\triangle ABC]=[\triangle ACD]+[\triangle BCD].</math> This can be rewritten as <math>\displaystyle \frac12 a b \sin C=\frac12 a d \sin \frac C2 +\frac12 b d \sin \frac C2 .</math><br />
<br />
Because <math>\displaystyle \sin C=2\sin \frac C2 \cos \frac C2,</math> the expression can be written as <math>\displaystyle2ab\cos \frac C2=d(a+b).</math> Dividing by <math>\displaystyle a+b,</math> <math>CD=\frac{2ab\cos \frac C2}{a+b},</math> as desired.<br />
<br />
----<br />
* [[1969 Canadian MO Problems/Problem 4|Previous Problem]]<br />
* [[1969 Canadian MO Problems/Problem 6|Next Problem]]<br />
* [[1969 Canadian MO Problems|Back to Exam]]</div>
4everwise